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Chapter 23: Problem 43

Three randomly chosen nonnegative integers \(x, y\) and \(z\) are found to satisfy the equation \(x+y+z=10\). Then the probability that \(z\) is even, is [A] \(\frac{36}{55}\) [B] \(\frac{6}{11}\) \([\mathrm{C}] \frac{1}{2}\) [D] \(\frac{5}{11}\)

Short Answer

Expert verified
\(\frac{6}{11}\)

Step by step solution

01

Identify the Total Possible Outcomes

The total number of ways to distribute the sum of 10 among three nonnegative integers \(x, y, z\) is equivalent to the number of solutions to the equation \(x+y+z=10\). This is a problem of distributing 10 identical objects into 3 distinct boxes, which can be calculated using stars and bars method. The formula is \(\binom{n+k-1}{k-1}\), where \(n\) is the number of objects and \(k\) is the number of boxes. So the total number of outcomes is \(\binom{10+3-1}{3-1}=\binom{12}{2}=66\).
02

Identify the Favorable Outcomes

Now we need to find the number of solutions where \(z\) is even. \((z=0,2,4,6,8,10)\). For each even value of \(z\), we can find the number of solutions to the equation \(x+y=10-z\) using stars and bars method. We add these numbers to find the total number of favorable outcomes.\For \(z=0\): \(x+y=10\), there are \(\binom{10+2-1}{2-1}=11\) solutions.\For \(z=2\): \(x+y=8\), there are \(\binom{8+2-1}{2-1}=9\) solutions.\For \(z=4\): \(x+y=6\), there are \(\binom{6+2-1}{2-1}=7\) solutions.\For \(z=6\): \(x+y=4\), there are \(\binom{4+2-1}{2-1}=5\) solutions.\For \(z=8\): \(x+y=2\), there are \(\binom{2+2-1}{2-1}=3\) solutions.\For \(z=10\): \(x+y=0\), there is \(1\) solution.\Adding these numbers gives us the total favorable outcomes: \(11+9+7+5+3+1=36\).
03

Calculate the Probability

The probability that \(z\) is even is the number of favorable outcomes divided by the total number of outcomes: \(\frac{36}{66} = \frac{6}{11}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorics
Combinatorics is a field of mathematics that deals with counting, arrangement, and combination of objects. It is foundational to several branches of mathematics and has practical applications in computer science, probability, and statistics. The most common problems involve finding the number of possible arrangements or selections from a set, which often relate to real-world scenarios such as determining probabilities.

For example, in the exercise mentioned, finding the number of nonnegative integer solutions to the equation x+y+z=10 is a combinatorial problem. Methods like stars and bars are often employed in combinatorial problems to simplify the counting process, especially when dealing with partitions of integers and nonnegative solutions. The focus here is not only on how many solutions there are, but also on the structure of these solutions—how they are composed and how each part contributes to the whole.
Stars and Bars Method
The stars and bars method is a combinatorial technique used to solve problems of distributing indistinguishable items (stars) into distinguishable bins (bars). It is particularly useful in problems where one needs to find the number of ways to distribute a certain number of identical objects among different recipients.

In the context of the given exercise, where we are finding the ways to allocate the sum of 10 among three variables x, y, z, stars and bars gives us a straightforward approach to arrive at the total number of nonnegative integer solutions. The unique aspect of this method is that it converts the problem into a scenario of placing dividers (bars) in between items (stars), such that each partition represents the count for one variable.
Nonnegative Integer Solutions
Nonnegative integer solutions refer to the solutions of an equation where all the variable values are integers greater than or equal to zero. Such constraints are common in problems involving counting or partitioning objects, where negative amounts do not make sense.

In the current problem, we require that x, y, z are nonnegative integers satisfying x+y+z=10. When solving for the number of nonnegative integer solutions using the stars and bars method, each solution represents a specific distribution of 10 (the sum) among the three variables. Every solution is a unique breakdown of the number 10 into three nonnegative parts, and calculating these solutions falls under the domain of combinatorial mathematics. The restriction to nonnegative integers is crucial; it simplifies the problem and connects directly with real-world scenarios such as dividing resources or tasks.
JEE Advanced Mathematics
JEE Advanced is an annual entrance examination in India for students seeking admission to various engineering colleges. It is known for its challenging mathematics section, which includes topics such as algebra, calculus, and indeed, combinatorics. A deep understanding of concepts like probability distribution, the stars and bars method, and combinatorial counting is essential for success in the exam.

Students preparing for the JEE Advanced Mathematics portion will encounter exercises like the textbook problem discussed here. These problems test the ability to apply theoretical concepts to practical problems and require an analytical approach to solving. Understanding the underlying principles and methods, such as those involved in finding nonnegative integer solutions, is key to excelling in this rigorous examination.

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Most popular questions from this chapter

A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is \(\delta T=0.01\) seconds and he measures the depth of the well to be \(L=20\) meters. Take the acceleration due to gravity \(g=10 \mathrm{~ms}^{-2}\) and the velocity of sound is \(300 \mathrm{~ms}^{-1}\). Then the fractional error in the measurement, \(\delta L / L\), is closest to [A] \(0.2 \%\) [B] \(1 \%\) [C] \(3 \%\) [D] \(5 \%\)

A photoelectric material having work-function \(\phi_{0}\) is illuminated with light of wavelength \(\lambda\left(\lambda<\frac{h c}{\phi_{0}}\right) .\) The fastest photoelectron has a de Broglie wavelength \(\lambda_{d} .\) A change in wavelength of the incident light by \(\Delta \lambda\) results in a change \(\Delta \lambda_{d}\) in \(\lambda_{d}\). Then the ratio \(\Delta \lambda_{d} / \Delta \lambda\) is proportional to \([\mathrm{A}] \quad \lambda_{d} / \lambda\) [B] \(\lambda_{d}^{2} / \lambda^{2}\) [C] \(\lambda_{d}^{3} / \lambda\) [D] \(\lambda_{d}^{3} / \lambda^{2}\)

A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the Sun and the Earth. The Sun is \(3 \times 10^{5}\) times heavier than the Earth and is at a distance \(2.5 \times 10^{4}\) times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is \(v_{e}=11.2 \mathrm{~km} \mathrm{~s}^{-1} .\) The minimum initial velocity \(\left(v_{S}\right)\) required for the rocket to be able to leave the Sun-Earth system is closest to (Ignore the rotation and revolution of the Earth and the presence of any other planet) [A] \(v_{S}=22 \mathrm{~km} \mathrm{~s}^{-1}\) [B] \(v_{S}=42 \mathrm{~km} \mathrm{~s}^{-1}\) [C] \(v_{S}=62 \mathrm{~km} \mathrm{~s}^{-1}\) [D] \(v_{S}=72 \mathrm{~km} \mathrm{~s}^{-1}\)

The order of the oxidation state of the phosphorus atom in \(\mathrm{H}_{3} \mathrm{PO}_{2}, \mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{3} \mathrm{PO}_{3}\), and \(\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}\) is [A] \(\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{3} \mathrm{PO}_{2}>\mathrm{H}_{3} \mathrm{PO}_{4}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}\) [B] \(\mathrm{H}_{3} \mathrm{PO}_{4}>\mathrm{H}_{3} \mathrm{PO}_{2}>\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}\) [C] \(\mathrm{H}_{3} \mathrm{PO}_{4}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}>\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{3} \mathrm{PO}_{2}\) [D] \(\mathrm{H}_{3} \mathrm{PO}_{2}>\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}>\mathrm{H}_{3} \mathrm{PO}_{4}\)

The instantaneous voltages at three terminals marked \(X, Y\) and \(Z\) are given by $$ \begin{aligned} &V_{x}=V_{0} \sin \omega t \\ &V_{Y}=V_{0} \sin \left(\omega t+\frac{2 \pi}{3}\right) \text { and } \\ &V_{z}=V_{0} \sin \left(\omega t+\frac{4 \pi}{3}\right) . \end{aligned} $$ An ideal voltmeter is configured to read \(r m s\) value of the potential difference between its terminals. It is connected between points \(X\) and \(Y\) and then between \(Y\) and \(Z\). The reading(s) of the voltmeter will be [A] \(\quad V_{X Y}^{r m s}=V_{0} \sqrt{\frac{3}{2}}\) [B] \(\quad V_{Y Z}^{r m s}=V_{0} \sqrt{\frac{1}{2}}\) [C] \(V_{X Y}^{r m s}=V_{0}\) [D] independent of the choice of the two terminals
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